Question

Find:
$\underset{x\to 0}{lim}\frac{1-\cos 2x}{{\sin }^{3}2x}$

Answer 1

$\underset{x\to 0}{lt}\frac{1-\cos 2x}{{\sin }^{3}2x}$

$=\underset{x\to 0}{lt}\frac{2{\sin }^{2}x}{(2\sin x\cos x{)}^3}$

$=\underset{x\to 0}{lt}\frac{2{\sin }^{2}x}{8{\sin }^{3}x{\cos }^{3}x}$

$=\underset{x\to 0}{lt}\frac{1}{4\sin x{\cos }^{3}x}=\frac{1}{\mathrm{4.0.1}}=\infty$.